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Chapter 3: Problem 17

In Problems 15-24, find the values of \(x \in\) R for which the given functions are both defined and continuous. $$ f(x)=\frac{x^{2}+1}{x-1} $$

Short Answer

Expert verified
The function is defined and continuous for \( x \in \mathbb{R} \setminus \{1\} \).

Step by step solution

01

Identify undefined points

To determine where the function \( f(x) = \frac{x^2 + 1}{x - 1} \) is undefined, set the denominator equal to zero and solve for \( x \). The denominator is zero when \( x - 1 = 0 \), which gives \( x = 1 \). Thus, the function is undefined at \( x = 1 \).
02

Determine continuity

A rational function like \( f(x) = \frac{x^2 + 1}{x - 1} \) is continuous at all points except where it is undefined. Since we found in Step 1 that the function is undefined at \( x = 1 \), the function is discontinuous at this point. Otherwise, it is continuous for all other real values of \( x \).
03

Solution summary

The function \( f(x) = \frac{x^2 + 1}{x - 1} \) is both defined and continuous for all real numbers \( x \) except \( x = 1 \). This means \( x \in \mathbb{R} \setminus \{1\} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Undefined Points in Functions
When working with mathematical functions, it's important to figure out where a function may be undefined. A function can become undefined when you encounter certain conditions that make its computation impossible or undefined. For instance, in the function \( f(x) = \frac{x^2 + 1}{x - 1} \), setting up the denominator \( x - 1 \) equal to zero can lead us to these points. Solving for \( x \), you find that at \( x = 1 \) the denominator equals zero, making the function undefined at this point.
  • Undefined points arise where the denominator of a rational function turns zero.
  • Once you find these points, the function cannot evaluate at these values of \( x \).
Recognizing undefined points is crucial because they directly influence the continuity and real-world applicability of a function.
Discontinuity
Discontinuity refers to a point or collection of points in the domain of a function where the function is not continuous. For rational functions, these discontinuities typically occur at the undefined points. In our function \( f(x) = \frac{x^2 + 1}{x - 1} \), the only point where it is discontinuous is \( x = 1 \).
  • At \( x = 1 \), the function jumps from being defined to undefined, thus creating a break in the continuity.
  • Beyond this point, the function continues without interruption over other real values of \( x \).
Identifying discontinuity helps in understanding where the behavior of the function changes abruptly. Knowing these points is essential if you're plotting graphs or interpreting function behavior.
Rational Functions
Rational functions are mathematical expressions represented as a ratio of two polynomials. In the expression \( f(x) = \frac{x^2 + 1}{x - 1} \), the numerator is \( x^2 + 1 \) and the denominator is \( x - 1 \). These functions have some intriguing characteristics due to their form:
  • They are continuous across most of their domain, starkly excepted where the denominator zeroes out.
  • The denominator can create vertical asymptotes, creating breaks in the graph where the function shoots off towards infinity.
  • The degree of polynomials affects the behavior 鈥� for instance, more complex polynomials might introduce additional asymptotes or undefined points.
Rational functions are prevalent in fields like engineering, physics, and economics, which use them to model real-world behaviors such as growths or decays, making an understanding of their properties invaluable.

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Most popular questions from this chapter

Use the Intermediate Value Theorem to show that \(\cos x=x\) has a solution in \((0,1)\).

In Problems 15-24, find the values of \(x \in\) R for which the given functions are both defined and continuous. $$ f(x)=3 x^{4}-x^{2}+4 $$

In Problems 29-48, find the limits. $$ \lim _{x \rightarrow-1} e^{x^{2} / 2-1} $$

(a) Show that $$ f(x)=\sqrt{x^{2}-4}, \quad|x| \geq 2 $$ is continuous from the right at \(x=2\) and continuous from the left at \(x=-2\). (b) Graph \(f(x)\). (c) Does it make sense to look at continuity from the left at \(x=2\) and at continuity from the right at \(x=-2 ?\)

The Hill equation is used to model how hemoglobin in blood binds to oxygen. If the proportion of hemoglobin molecules that are bound to oxygen is \(h\) and the concentration of oxygen (measured as a partial pressure, that varies from 0 to \(\infty\) ) is \(P\), then a common model is: $$h(P)=\frac{a P^{k}}{30^{k}+P^{k}}$$ where \(k \geq 1\) and \(a>0\) are constants that depend on the species of animal and its environment (e.g., whether it lives at sea-level or at altitude). (a) Show that no matter what the values of \(a\) and \(k\) are, the amount of bound oxygen goes to zero as the oxygen concentration goes to \(0 ;\) that is: $$\lim _{P \rightarrow 0} h(P)=0$$ (b) It is known that as \(P\) increases, the amount of bound oxygen plateaus. Since \(h=1\) when all hemoglobin molecules are bound to oxygen, we want our model to reflect that: $$\lim _{P \rightarrow \infty} h(P)=1$$ This is called the saturation value for oxygen binding. Explain what value of \(a\) must be chosen for this condition to be satisfied. (c) The half-saturation constant, \(P_{1 / 2}\), is defined to be the concentration of oxygen at which the proportion of bound hemoglobin molecules reaches half its saturation value, that is: $$h\left(P_{1 / 2}\right)=\frac{1}{2} \lim _{P \rightarrow \infty} h(P)$$ Show that \(P_{1 / 2}=30\). (d) In a patient with carbon monoxide poisoning carbon monoxide binds preferentially to the hemoglobin instead of oxygen, stopping the blood from effectively transporting oxygen around the body. For a patient with acute carbon monoxide poisoning, the relationship between proportion of bound hemoglobin molecules and oxygen concentration can be modeled by: \(h(P)=\frac{0.9 P^{3}}{P^{3}+26^{3}} \quad\) (we have assumed that \(k=3\) ) Show that both the saturation level for oxygen binding and the half-saturation constant are both changed from your answers in (b) and (c).
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